Cumulative frequency — ogive, median from graph: Worked examples

2.7.2 · D3 · Maths › Statistics & Probability — Intermediate › Cumulative frequency — ogive, median from graph

Scenario matrix

Har median-from-CF problem in cells mein se ek (ya blend) hoti hai:

Cell	Kya special hai	Kaun sa trap set karta hai
A	$N /2$ exactly ek class boundary par land karta hai (CF equals $N /2$ )	Off-by-one median class
B	$N /2$ mid-class land karta hai — real interpolation	Rounding / fraction slips
C	Discontinuous classes ( $0 - 9, 10 - 19$ )	$\pm 0.5$ boundary fix bhool jaana
D	Twin-ogive — crossing se median padhna	Galat axis padhna
E	Unequal class widths ( $h$ har class mein alag)	Galat $h$ use karna
F	Missing frequency given the median	Do equations saath solve karna
G	Degenerate: median pehli/aakhiri class mein, open-ended interval, saara data ek class mein	$C F_{b} = 0$ ; open-end boundaries
H	Real-world word problem (waiting times, incomes)	Pehle table extract karo

Ab hum har cell ko kam se kam ek baar 8 worked examples mein cover karte hain (Cell G ko G(i) aur G(ii) mein split kiya hai).

Setup jo hum baar baar use karte hain

Example 1 — Cell A · $N /2$ boundary par

Worked example Cell A

40 students ke marks. Median nikalo.

Marks	$f$	CF
0–10	4	4
10–20	6	10
20–30	10	20
30–40	12	32
40–50	8	40

Forecast: $N /2 = 20$ . CF exactly 20 par "less than 30" mein hit karta hai. Kya median class 20–30 hai ya 30–40? Aage padhne se pehle guess karo.

$N /2 = 40/2 = 20$ . Yeh step kyun? Grouped data mein hum height $N /2$ se split karte hain, kabhi $(N + 1) /2$ se nahi — ogive ek smooth curve hai, discrete list nahi.
Median class dhundho. CF $10$ (below 20) se $20$ tak 20–30 class ke andar jaata hai. Toh median class $= 20 - 30$ . Yeh step kyun? Median class wahi hai jahan CF neeche se $N /2$ cross kare ya reach kare — CF $x = 20$ par $10 < 20$ tha, phir is class ke across $20$ tak chada. Yeh 30–40 NAHI hai.
Extract karo: $L = 20, C F_{b} = 10, f = 10, h = 10$ . Yeh step kyun? $C F_{b}$ 20–30 se pehle wali class ka CF hai (woh 10–20 row hai, CF $= 10$ ).
Interpolate karo: $Median = 20 + (\frac{20 - 10}{10}) \times 10 = 20 + 1 \times 10 = 30.$ Yeh step kyun? Humhe $20 - 10 = 10$ aur counts chahiye the; class mein exactly $f = 10$ hain, toh hum poori width travel karte hain — far edge par land karte hain, $x = 30$ .

Figure — Cumulative frequency — ogive, median from graph

Example 2 — Cell B · interpolation genuinely kaam karta hai

$N /2 = 20$ . Kyun? Wohi halving rule.
Median class. CF 20–30 ke andar $13 \to 25$ jump karta hai, toh 20th count usme hai. Median class $= 20 - 30$ . Kyun? $13 < 20 \leq 25$ — crossing is class ke andar hoti hai.
Extract karo: $L = 20, C F_{b} = 13, f = 12, h = 10$ .
Interpolate karo: $Median = 20 + (\frac{20 - 13}{12}) \times 10 = 20 + \frac{7}{12} \times 10 = 20 + 5.8 \overline{3} = 25.83.$ Kyun? Hum class mein $7$ counts andar hain, class mein $12$ hain; maante hain ki 12 values width 10 mein evenly spread hain, toh hum $\frac{7}{12}$ way andar jaate hain. Convention ke mutabiq final line ko $25.83$ round karte hain.

Example 3 — Cell C · discontinuous classes

Worked example Cell C

50 logon ki ages (whole years). Classes discontinuous hain.

Age	$f$
10–19	6
20–29	14
30–39	18
40–49	8
50–59	4

Forecast: 19 aur 20 ke beech ka gap aadha saal chhupa raha hai. Agar fix nahi kiya, median $0.5$ se off ho jaayega. Pehle corrected boundaries guess karo.

Continuous boundaries mein convert karo. Har lower se $0.5$ ghataao, har upper mein $0.5$ jodao: $9.5 - 19.5, 19.5 - 29.5, 29.5 - 39.5, \dots$ Yeh step kyun? 19 aur 20 ke beech real ages exist karti hain; true boundary midpoint $19.5$ hai. Ab $h = 10$ cleanly milta hai aur koi gaps nahi.
CF table: $6, 20, 38, 46, 50$ .
$N /2 = 25$ . CF $29.5 - 39.5$ class mein $20 \to 38$ jaate hue $25$ cross karta hai. Median class $= 29.5 - 39.5$ .
Extract karo: $L = 29.5, C F_{b} = 20, f = 18, h = 10$ .
Interpolate karo: $Median = 29.5 + (\frac{25 - 20}{18}) \times 10 = 29.5 + \frac{5}{18} \times 10 = 29.5 + 2.78 = 32.28.$ Kyun? $5$ counts class of $18$ mein, width 10; do decimals tak round kiya.

Example 4 — Cell D · twin-ogive (crossing padhna)

Forecast: Dono curves $y = N /2 = 20$ par cross karte hain. Guess: wahan $x$ ko Example 2 ke $25.83$ se match karna chahiye.

Less-than CF: $5, 13, 25, 35, 40$ ko upper boundaries $10, 20, 30, 40, 50$ par plot karo. Kyun upper? Less-than CF upper boundary se neeche sab count karta hai.
More-than CF: $40, 35, 27, 15, 5$ ko lower boundaries $0, 10, 20, 30, 40$ par plot karo. Kyun lower? Lower boundary par more-than CF $=$ us boundary se right tak sab kuch.
Yeh cross karte hain jahan "below ka number" $=$ "above ka number" $= N /2 = 20$ . Perpendicular giraao → $x \approx 25.8$ . Yeh step kyun? Dono halves $N /2$ ke barabar sirf median par hote hain — median ki definition se.

Example 5 — Cell E · unequal class widths

Worked example Cell E

$N = 60$ . Widths alag hain — $h$ har class ke liye padho.

Class	width $h$	$f$	CF
0–10	10	8	8
10–30	20	20	28
30–40	10	22	50
40–70	30	10	60

Forecast: $N /2 = 30$ . Kaunsi class mein hai, aur us class ki apni width kya hai?

$N /2 = 30$ . CF 30–40 class mein $28 \to 50$ jaate hue $30$ cross karta hai.
IS class ki width use karke extract karo: $L = 30, C F_{b} = 28, f = 22, h = 10$ . Yeh step kyun? Formula ka $h$ sirf median class ki width hai — yahan $10$ , chahe pehle wali class ki width $20$ thi. Neighbour ki width use karna hi Cell-E trap hai.
Interpolate karo: $Median = 30 + (\frac{30 - 28}{22}) \times 10 = 30 + \frac{2}{22} \times 10 = 30 + 0.909 = 30.91.$ Kyun? $2$ counts class of $22$ mein, width 10; do decimals tak round kiya.

Example 6 — Cell F · missing frequency solve karo (honest algebra)

Worked example Cell F

$N = 100$ aur median known hai $35$ . 30–40 class ki frequency $x$ missing hai. $x$ nikalo.

Class	$f$
0–10	8
10–20	12
20–30	20
30–40	$x$
40–50	18
50–60	$y$

Forecast: Do unknowns ( $x$ aur $y$ ), toh do equations chahiye: total $= 100$ aur median formula $= 35$ . Inhe saath solve karo. Guess karo ki $x$ whole number aayega ya nahi.

Equation 1 — total constraint. $8 + 12 + 20 + x + 18 + y = 100 \Rightarrow x + y = 42.$ Yeh step kyun? Saari frequencies $N$ mein sum honi chahiye; do blanks ke saath yeh do unknowns mein ek equation hai, toh akele isse finish nahi kar sakte — median equation bhi chahiye.
Given median se median class locate karo. Stated median $35 \in (30, 40)$ hai, toh median class 30–40 hai. Yeh step kyun? Median hamesha apni class ke andar hota hai; $35$ batana class ko $35$ ke around pin kar deta hai.
Paanch numbers extract karo (median class 30–40). $L = 30, C F_{b} = 8 + 12 + 20 = 40, f = x, h = 10, N = 100.$ Yeh step kyun? $C F_{b}$ 30–40 se pehle ki teen classes ka running total hai; $f$ median class ki khud ki unknown frequency hai. Consistency check: median ke 30–40 ke andar hone ke liye $C F_{b} < N /2 = 50$ chahiye; yahan $40 < 50$ ✓, toh problem ka solution exist karta hai — koi "misprint" surgery nahi chahiye.
Equation 2 — median formula. Ek formula mein substitute karo aur $35$ ke barabar set karo: $35 = 30 + (\frac{\frac{100}{2} - 40}{x}) \times 10 \Rightarrow 5 = \frac{50 - 40}{x} \times 10 = \frac{100}{x} .$ Yeh step kyun? $x$ ke alaawa sab known hai, toh median equation $x$ mein ek equation ban jaati hai.
$x$ solve karo. $5 = \frac{100}{x} \Rightarrow x = \frac{100}{5} = 20.$ Yeh step kyun? Dono sides ko $x$ se multiply karo, phir $5$ se divide — pure algebra, clean whole number.
Equation 1 use karke finish karo. $x + y = 42$ aur $x = 20$ , toh $y = 42 - 20 = 22.$ Yeh step kyun? Total constraint doosra unknown deliver karta hai jab median pehle ko pin kar de.

Example 7 — Cell G · degenerate cases

Worked example Cell G(i) — median PEHLI class mein

$N = 20$ . Data low end par pile up hua hai.

Class	$f$	CF
0–5	12	12
5–10	5	17
10–15	3	20

Forecast: $N /2 = 10$ . CF pehli class ke andar hi $10$ reach kar leta hai. Jab pehle koi class na ho toh $C F_{b}$ kya hoga?

$N /2 = 10$ . CF 0–5 ke andar $10$ cross karta hai (CF $0 \to 12$ jaata hai). Median class $= 0 - 5$ .
Extract karo: $L = 0, C F_{b} = 0, f = 12, h = 5$ . $C F_{b} = 0$ kyun? Pehle koi class nahi hai, toh kuch accumulate nahi hua — running total $0$ se shuru hota hai. Yahi degenerate boundary case hai.
Interpolate karo: $Median = 0 + (\frac{10 - 0}{12}) \times 5 = \frac{10}{12} \times 5 = 4.17.$ Yeh step kyun? Hume is class ke $f = 12$ mein se $10$ counts chahiye; woh $12$ values width $5$ mein evenly spread hain, toh hum width ka $\frac{10}{12}$ travel karte hain. $L = 0$ ke saath poora median wahi step hai.

Worked example Cell G(ii) — OPEN-ENDED interval ("60+") aur LAST class

50 logon ki monthly incomes (₹ thousand). Top class ki koi upper bound nahi.

Income	$f$	CF
0–20	8	8
20–40	20	28
40–60	14	42
60+	8	50

Forecast: "60+" ki koi upper edge nahi — kya yeh median bigaad dega? Guess: kya open end yahan matter bhi karta hai?

$N /2 = 25$ . CF 20–40 class mein $8 \to 28$ jaate hue $25$ cross karta hai. Median class $= 20 - 40$ . Yeh step kyun? $8 < 25 \leq 28$ — crossing 20–40 ke andar hai, open top class ke paas kahin nahi.
Extract karo: $L = 20, C F_{b} = 8, f = 20, h = 20$ . Yeh step kyun? Paanchon numbers mein se har ek closed class mein hai (20–40 aur uska predecessor), toh "60+" ki missing upper edge formula mein kabhi nahi aati.
Interpolate karo: $Median = 20 + (\frac{25 - 8}{20}) \times 20 = 20 + \frac{17}{20} \times 20 = 20 + 17 = 37.$ Yeh step kyun? $17$ counts class of $20$ mein, width $20$ .

Example 8 — Cell H · real-world word problem

Worked example Cell H

Ek café 80 customers ke wait time (minutes) record karta hai. "Aadhe customers kitne der se kam waited?" — yahi median hai.

Wait (min)	$f$
0–2	10
2–4	22
4–6	28
6–8	14
8–10	6

Forecast: "Aadhe waited less than…" seedha median chilla raha hai. $N /2 = 40$ . Guess karo kaun sa 2-minute band.

Words translate karo. "Woh time jiske neeche aadhe customers fall karein" $=$ median. Yeh step kyun? Word problems table chhupate hain; "aadhe below fall" phrase median ki definition hai.
CF: $10, 32, 60, 74, 80$ . $N /2 = 40$ .
Median class: CF 4–6 mein $32 \to 60$ jaate hue $40$ cross karta hai. Extract karo $L = 4, C F_{b} = 32, f = 28, h = 2$ .
Interpolate karo: $Median = 4 + (\frac{40 - 32}{28}) \times 2 = 4 + \frac{8}{28} \times 2 = 4 + 0.571 = 4.57 min .$ Kyun? $8$ counts class of $28$ mein, width $2$ ; do decimals tak round kiya.

First class median ::: $C F_{b} = 0$ (abhi tak kuch accumulate nahi hua) Discontinuous classes ::: $L, h$ dhundhne se pehle $0.5$ subtract/add karo $N /2$ boundary par ::: median class woh hai jo CF neeche se cross kare, agle wali nahi Unequal widths ::: median class ka apna $h$ use karo Missing frequency + given median ::: total equation AUR median equation likho, saath solve karo Missing frequency ka koi solution nahi jab ::: $C F_{b} \geq N /2$ (pehle ki classes already halfway pahunch chuki hain) Open-ended interval ::: harmless hai agar median class closed ho; warna neighbour ka $h$ assume karo Twin ogive ::: median woh $x$ hai jahan dono curves height $N /2$ par cross karti hain

Worked examples — Cumulative frequency — ogive, median from graph